Polynomial and Rational Inequalities Questions with Solutions

Explore a variety of multiple-choice questions focused on polynomial and rational inequalities, complete with detailed solutions and step-by-step explanations to help deepen your understanding. If you're new to the topic or want a refresher, consider starting with this comprehensive tutorial on inequalities before tackling the problems.

Question 1

Solve the inequality \[ -(x + 2) + 2x > 2(x - 3) + 3x \]

  1. \(\text{No solutions}\)
  2. \((-\infty, \dfrac{4}{5})\)
  3. \((\dfrac{4}{5}, +\infty)\)
  4. \((-\infty, +\infty)\)
  5. \((-\infty, 1)\)

Solution:

  • Given inequality:
    \[-(x + 2) + 2x > 2(x - 3) + 3x\]
  • Expand both sides:
    Left side:
    \[-(x + 2) + 2x = -x - 2 + 2x = x - 2\]
    Right side:
    \[2(x - 3) + 3x = 2x - 6 + 3x = 5x - 6\]
  • Simplified inequality:
    \[x - 2 > 5x - 6\]
  • Move variables and constants:
    \[x - 5x - 2 > -6\]
    \[-4x - 2 > -6\]
    \[-4x > -4\]
  • Divide both sides by \(-4\), and reverse the inequality sign:
    \[x < \dfrac{-4}{-4} = 1\]
  • Final answer in interval notation:
    \[\boxed{(-\infty, 1)}\]

Question 2

Solve the polynomial inequality \[ (6x + 2) (10 - 20x) \ge 0 \]

  1. \( \left( -\dfrac{1}{3}, \dfrac{1}{2} \right) \)
  2. \( \left[ -\dfrac{1}{2}, +\infty \right) \)
  3. \( \text{No solutions} \)
  4. \( \left( -\infty, 0 \right] \)
  5. \( \left[ -\dfrac{1}{3}, \dfrac{1}{2} \right] \)

Solution:

  • Given inequality:
    \[(6x + 2)(10 - 20x) \ge 0\]
  • Find the roots (zeros) of the expression:
    Set each factor equal to zero:
    \[ 6x + 2 = 0 \Rightarrow x = -\dfrac{1}{3} \]
    \[ 10 - 20x = 0 \Rightarrow x = \dfrac{1}{2} \]
  • The expression changes sign at the critical points:
    \[x = -\dfrac{1}{3}, \quad x = \dfrac{1}{2}\]
  • Test the sign of the expression in each interval:
    • Interval: \(x < -\dfrac{1}{3}\), e.g., \(x = -1\)
      \((6(-1) + 2)(10 - 20(-1)) = (-6 + 2)(10 + 20) = (-4)(30) = -120\) ? negative
    • Interval: \(-\dfrac{1}{3} < x < \dfrac{1}{2}\), e.g., \(x = 0\)
      \((6(0) + 2)(10 - 20(0)) = (2)(10) = 20\) ? positive
    • Interval: \(x > \dfrac{1}{2}\), e.g., \(x = 1\)
      \((6(1) + 2)(10 - 20(1)) = (6 + 2)(-10) = (8)(-10) = -80\) ? negative
  • Include endpoints because of "=" (greater than or equal to):
    \[ (6x + 2)(10 - 20x) \ge 0 \Rightarrow x \in \left[-\dfrac{1}{3}, \dfrac{1}{2}\right] \]
  • Final answer in interval notation:
    \[\boxed{\left[-\dfrac{1}{3}, \dfrac{1}{2}\right]}\]

Question 3

Solve the inequality: \[ x^2 + 3x - 4 > 0 \]

  1. \( (-4, 1) \)
  2. \( (-\infty, -4) \cup (1, +\infty) \)
  3. \( (1, +\infty) \)
  4. \( \text{No solutions} \)
  5. \( (-\infty, -4] \cup [1, +\infty) \)

Solution:

  • Given inequality:
    \[x^2 + 3x - 4 > 0\]
  • Factor the quadratic expression:
    \[x^2 + 3x - 4 = (x + 4)(x - 1)\]
  • Rewrite the inequality:
    \[(x + 4)(x - 1) > 0\]
  • Find the critical points by setting each factor to 0:
    \[x + 4 = 0 \Rightarrow x = -4\]
    \[x - 1 = 0 \Rightarrow x = 1\]
  • Test sign of the expression in each interval defined by the critical points:
    • Interval \((-\infty, -4)\): pick \(x = -5\)
      \[(x + 4)(x - 1) = (-1)(-6) = 6 > 0\]
    • Interval \((-4, 1)\): pick \(x = 0\)
      \[(x + 4)(x - 1) = (4)(-1) = -4 < 0\]
    • Interval \((1, \infty)\): pick \(x = 2\)
      \[(x + 4)(x - 1) = (6)(1) = 6 > 0\]
  • Determine where the expression is positive:
    The expression is positive in:
    \((-\infty, -4)\) and \((1, \infty)\)
  • Final answer in interval notation:
    \[\boxed{(-\infty, -4) \cup (1, \infty)}\]

Question 4

Solve the inequality \[ 2x^2 \lt 32 \]

  1. \( (-4,\, 4) \)
  2. \( (-\infty,\, 4) \)
  3. \( \left(-\sqrt{32},\, \sqrt{32} \right) \)
  4. No solutions
  5. \( (-4,\, +\infty) \)

Solution:

  • Given inequality:
    \[2x^2 < 32\]
  • Divide both sides by 2:
    \[\dfrac{2x^2}{2} < \dfrac{32}{2}\]
    \[x^2 < 16\]
  • Take square roots of both sides (remember to consider both positive and negative roots):
    \[-4 < x < 4\]
  • Final answer in interval notation:
    \[\boxed{(-4,\ 4)}\]

Question 5

Solve the inequality \[ 25 x^2 \le 9x \]

  1. No solutions
  2. \( (-\infty , +\infty) \)
  3. \( [0 , +\infty) \)
  4. \( \left[0 , \dfrac{9}{25} \right] \)
  5. \( \left[0 , \dfrac{9}{5} \right) \)

Solution:

  • Given inequality:
    \[25x^2 \le 9x\]
  • Move all terms to one side to form a standard quadratic inequality:
    \[25x^2 - 9x \le 0\]
  • Factor the quadratic expression:
    \[x(25x - 9) \le 0\]
  • Find the critical points by solving:
    \[x = 0 \quad \text{or} \quad 25x - 9 = 0 \Rightarrow x = \dfrac{9}{25}\]
  • Test intervals between critical points:
    The critical points divide the number line into intervals:
    \[ \text{Interval 1: } (-\infty, 0) \\ \text{Interval 2: } (0, \dfrac{9}{25}) \\ \text{Interval 3: } \left(\dfrac{9}{25}, \infty\right) \] Choose test points in each interval:
    For \(x = -1\): \(x(25x - 9) = (-1)(-25 - 9) = 34 > 0\)
    For \(x = \dfrac{1}{10}\): \(\dfrac{1}{10}(25 \cdot \dfrac{1}{10} - 9) = \dfrac{1}{10}(-6.5) = -0.65 < 0\)
    For \(x = 1\): \(1(25 - 9) = 16 > 0\)
    So the inequality is satisfied only in the interval: \[ \left[0, \dfrac{9}{25}\right] \]
  • Final answer in interval notation:
    \[\boxed{[0, \dfrac{9}{25}]}\]

Question 6

Solve the inequality \[ \dfrac{1 + x} {|x - 2|} \gt 0 \]

  1. \( (-1 , +\infty) \)
  2. \( (-1 , 2) \cup (2 , +\infty) \)
  3. \( [-1 , +\infty) \)
  4. \( (2 , +\infty) \)
  5. \( (-\infty , +\infty) \)

Solution:

  • Given inequality:
    \[\dfrac{1 + x}{|x - 2|} > 0\]
  • Note: The denominator \(|x - 2|\) is always positive except at \(x = 2\), where it is undefined.
    So, we only need to determine when the numerator \(1 + x > 0\), since the denominator is positive.
    Domain restriction: \(x \ne 2\)
  • Solve the numerator inequality:
    \[1 + x > 0 \quad \Rightarrow \quad x > -1\]
  • Conclusion: The expression is positive when:
    \[ \begin{cases} x > -1 \\ x \ne 2 \end{cases} \]
  • Final answer in interval notation:
    \[\boxed{(-1, 2) \cup (2, \infty)}\]

Question 7

Solve the inequality \[ \dfrac{(2x^{2} + 3)(x + 2)}{x^{2} - 4} \geq 0 \]

  1. \((-\infty, +\infty)\)
  2. \((-\infty, 2)\)
  3. \((2, +\infty)\)
  4. \([2, +\infty)\)
  5. \( \text{no solutions} \)

Solution:

  • Given inequality:
    \[ \dfrac{(2x^2 + 3)(x + 2)}{x^2 - 4} \geq 0 \]
  • Factor the denominator:
    \[ x^2 - 4 = (x - 2)(x + 2) \] So the inequality becomes:
    \[ \dfrac{(2x^2 + 3)(x + 2)}{(x - 2)(x + 2)} \geq 0 \]
  • Cancel common factors:
    \(x + 2\) appears in both numerator and denominator, but we **cannot cancel it completely** because division by zero is undefined.
    We must note that \(x \neq -2\), and simplify to:
    \[ \dfrac{2x^2 + 3}{x - 2} \geq 0,\quad \text{with } x \neq -2 \]
  • Analyze the expression:
    - The numerator \(2x^2 + 3\) is always positive for all real \(x\) since it has no real roots and opens upward.
    - The sign of the expression depends on the denominator \(x - 2\)
  • Determine where the expression is nonnegative:
    - When \(x - 2 > 0 \Rightarrow x > 2\), the whole expression is positive
    - When \(x - 2 < 0 \Rightarrow x < 2\), the whole expression is negative
    - At \(x = 2\), the expression is undefined
    - At \(x = -2\), the expression is also undefined due to the original denominator
  • Solution:
    The expression is nonnegative when:
    \[ \dfrac{2x^2 + 3}{x - 2} \geq 0 \Rightarrow x > 2 \] (we exclude \(x = -2\) and \(x = 2\) because they make the denominator zero)
  • Final answer in interval notation:
    \[ \boxed{(2, \infty)} \]

Question 8

Solve the inequality \[ \dfrac{4}{6x - 4} \leq \dfrac{2}{2x + 2} \]

  1. \((-\infty, -1] \cup [4, +\infty)\)
  2. \(\left(\dfrac{2}{3}, 4\right]\)
  3. \((-1, \dfrac{2}{3})\)
  4. \((-1, \dfrac{2}{3}) \cup [4, +\infty)\)
  5. \([4, +\infty)\)

Solution:

  • Given inequality:
    \[\dfrac{4}{6x - 4} \leq \dfrac{2}{2x + 2}\]
  • Simplify the denominators:
    Factor each:
    \[6x - 4 = 2(3x - 2), \quad 2x + 2 = 2(x + 1)\]
    So the inequality becomes:
    \[\dfrac{4}{2(3x - 2)} \leq \dfrac{2}{2(x + 1)}\]
  • Cancel the common factor of 2 in the denominators:
    \[\dfrac{2}{3x - 2} \leq \dfrac{1}{x + 1}\]
  • Bring all terms to one side:
    \[\dfrac{2}{3x - 2} - \dfrac{1}{x + 1} \leq 0\]
  • Find common denominator:
    \[ \dfrac{2(x + 1) - 1(3x - 2)}{(3x - 2)(x + 1)} \leq 0 \] Simplify numerator:
    \[ 2(x + 1) - 3x + 2 = 2x + 2 - 3x + 2 = -x + 4 \] So the inequality becomes:
    \[ \dfrac{-x + 4}{(3x - 2)(x + 1)} \leq 0 \]
  • Find critical points by setting numerator and denominator to 0:
    \[ -x + 4 = 0 \Rightarrow x = 4 \] \[ 3x - 2 = 0 \Rightarrow x = \dfrac{2}{3}, \quad x + 1 = 0 \Rightarrow x = -1 \]
  • Make sign chart around the critical points: \(x = -1\), \(\dfrac{2}{3}\), and \(4\)
    Test intervals:
    • \(x < -1\): pick \(x = -2\)
      \[ \dfrac{-(-2) + 4}{(3(-2) - 2)(-2 + 1)} = \dfrac{2 + 4}{(-6 - 2)(-1)} = \dfrac{6}{(-8)(-1)} = \dfrac{6}{8} > 0 \]
    • \(-1 < x < \dfrac{2}{3}\): pick \(x = 0\)
      \[ \dfrac{-0 + 4}{(0 - 2)(0 + 1)} = \dfrac{4}{(-2)(1)} = \dfrac{4}{-2} = -2 < 0 \]
    • \(\dfrac{2}{3} < x < 4\): pick \(x = 1\)
      \[ \dfrac{-1 + 4}{(3 - 2)(2)} = \dfrac{3}{(1)(2)} = \dfrac{3}{2} > 0 \]
    • \(x > 4\): pick \(x = 5\)
      \[ \dfrac{-5 + 4}{(15 - 2)(6)} = \dfrac{-1}{13 \cdot 6} < 0 \]
  • Determine intervals where the expression is = 0:
    This occurs where the expression is negative **or zero**:
    • Negative: in intervals \((-1, \dfrac{2}{3})\) and \((4, \infty)\)
    • Zero: when numerator is zero ? \(x = 4\)
    • Exclude points where denominator is zero: \(x = -1\), \(x = \dfrac{2}{3}\)
    So the solution is:
    \[ (-1, \dfrac{2}{3}) \cup [4, \infty) \]
  • Final answer in interval notation:
    \[\boxed{(-1, \dfrac{2}{3}) \cup [4, \infty)}\]

Question 9

Solve the double inequality. \[ 2 < \dfrac{|x + 1|}{2} - 2 < 5 \]

  1. : \((-3, -15)\)
  2. : no solutions
  3. : \((-15, -9) \cup (7, 13)\)
  4. : \((7, 13)\)
  5. : \((-\infty, +\infty)\)

Solution:

  • Given double inequality:
    \[2 < \dfrac{|x + 1|}{2} - 2 < 5\]
  • Add 2 to all parts of the inequality:
    \[2 + 2 < \dfrac{|x + 1|}{2} < 5 + 2\]
    \[4 < \dfrac{|x + 1|}{2} < 7\]
  • Multiply all parts by 2:
    \[2 \cdot 4 < |x + 1| < 2 \cdot 7\]
    \[8 < |x + 1| < 14\]
  • Now solve the compound absolute value inequality:
    \[8 < |x + 1| < 14\]
    This gives two separate inequalities:
    \[\text{Either } x + 1 < -8 \quad \text{or} \quad x + 1 > 8\]
    and
    \[x + 1 < 14 \quad \text{and} \quad x + 1 > -14\]
  • Now combine both conditions:
    \[\text{Valid values of } x \text{ satisfy:}\]
    \[(x + 1 < -8 \text{ and } x + 1 > -14) \quad \text{or} \quad (x + 1 > 8 \text{ and } x + 1 < 14)\]
  • Simplify each case:
    Case 1:
    \[-14 < x + 1 < -8 \Rightarrow -15 < x < -9\]
    Case 2:
    \[8 < x + 1 < 14 \Rightarrow 7 < x < 13\]
  • Final answer in interval notation:
    \[\boxed{(-15, -9) \cup (7, 13)}\]

Question 10

Solve the inequality \[ x^{2} < -x^{4} \]

  1. : no solutions
  2. : \((-1, 1)\)
  3. : \((-\infty, +\infty)\)
  4. : \((0, +\infty)\)
  5. : \([0, +\infty)\)

Solution:

  • Given inequality:
    \[ x^{2} < -x^{4} \]
  • Rewrite the inequality by bringing all terms to one side:
    \[ x^{2} + x^{4} < 0 \]
  • Factor the left side:
    \[ x^{2} + x^{4} = x^{2}(1 + x^{2}) \]
  • Analyze the factors:
    \[ x^{2} \geq 0 \quad \text{for all real } x \] and \[ 1 + x^{2} > 0 \quad \text{for all real } x \]
  • Since both factors are always \(\geq 0\) and \(> 0\) respectively, their product
    \[ x^{2}(1 + x^{2}) \geq 0 \] for all \(x\), so
    \[ x^{2}(1 + x^{2}) < 0 \] has no real solutions.
  • Final answer:
    \[ \boxed{\text{No real } x \text{ satisfy } x^{2} < -x^{4}} \]

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